Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-r + 9}{r^2 - 64} \div \dfrac{3r^2 + 27r}{r^3 + 17r^2 + 72r} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-r + 9}{r^2 - 64} \times \dfrac{r^3 + 17r^2 + 72r}{3r^2 + 27r} $ First factor out any common factors. $q = \dfrac{-(r - 9)}{r^2 - 64} \times \dfrac{r(r^2 + 17r + 72)}{3r(r + 9)} $ Then factor the quadratic expressions. $q = \dfrac {-(r - 9)} {(r + 8)(r - 8)} \times \dfrac {r(r + 8)(r + 9)} {3r(r + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-(r - 9) \times r(r + 8)(r + 9) } { (r + 8)(r - 8) \times 3r(r + 9)} $ $q = \dfrac {-r(r + 8)(r + 9)(r - 9)} {3r(r + 8)(r - 8)(r + 9)} $ Notice that $(r + 8)$ and $(r + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-r\cancel{(r + 8)}(r + 9)(r - 9)} {3r\cancel{(r + 8)}(r - 8)(r + 9)} $ We are dividing by $r + 8$ , so $r + 8 \neq 0$ Therefore, $r \neq -8$ $q = \dfrac {-r\cancel{(r + 8)}\cancel{(r + 9)}(r - 9)} {3r\cancel{(r + 8)}(r - 8)\cancel{(r + 9)}} $ We are dividing by $r + 9$ , so $r + 9 \neq 0$ Therefore, $r \neq -9$ $q = \dfrac {-r(r - 9)} {3r(r - 8)} $ $ q = \dfrac{-(r - 9)}{3(r - 8)}; r \neq -8; r \neq -9 $